A bag contains 1 red, 7 black and two yellow marbles. What is the probability of selecting 1 yellow marble, not replacing it then selecting another yellow marble?
How do you solve this?
there are initially 2 yellows out of a total of 10 marbles.
chance of yellow on 1st draw: 2/10
with that yellow gone, there is now only 1 yellow out of 9 total, so the chance of a yellow on the 2nd draw is 1/9
So, the total probability is 2/10 * 1/9
To solve this problem, you need to understand the concept of probability. Probability is the likelihood of an event occurring.
First, let's determine the total number of marbles in the bag. The bag contains 1 red marble, 7 black marbles, and 2 yellow marbles. Therefore, there are a total of 1 + 7 + 2 = 10 marbles in the bag.
Next, let's calculate the probability of selecting the first yellow marble. Since there are 2 yellow marbles out of a total of 10 marbles, the probability of selecting a yellow marble on the first draw is 2/10, which simplifies to 1/5.
After the first draw, we need to consider that we have not replaced the yellow marble back into the bag. Therefore, there is now one less marble in the bag. Thus, the total number of remaining marbles is 9.
Now, let's calculate the probability of selecting another yellow marble on the second draw. Since we have not replaced the first yellow marble, there is now 1 yellow marble left in the bag. So the probability of selecting a yellow marble on the second draw is 1/9.
Finally, to find the probability of both events happening, we multiply the probabilities of each event together. So the probability of selecting one yellow marble, not replacing it, and then selecting another yellow marble is (1/5) * (1/9), which simplifies to 1/45.
Therefore, the probability of selecting 1 yellow marble, not replacing it, and then selecting another yellow marble is 1/45.